Sample Task Kit for Home Lending

Answers: Secondary Kit

Look for your task name in alphabetical
order, or use Control F to search.

A Rectangle Of Squares

1. The rectangle has a total area of 1,056 sq. cm.
2. The rectangle is 33cm x 32cm.Extension
We do not know of any other set of squares which create a rectangle. Nor can we prove that there is no other. You might be able to find another, or, you might be able to prove that there cannot be another set.

A Stacking Problem
1. The puzzle takes 60 moves as follows:

Action

Moves

Total So Far

Blocks 1 & 2 to Position C.

3

Blocks 3 to Position B.

1

Blocks 1 & 2 back onto Block 3.

3

7 moves

Block 4 to Position C.

1

Blocks 1 - 3 onto Block 4.

7

15 moves

Block 5 to Position B.

1

Blocks 1 - 4 onto Block 5.

15

31 moves

Block 6 to Position C.

1

Blocks 1 - 4 to Position A.

15

47 moves

Block 5 onto Block 6.

1

Blocks 1- 3 to Position C.

7

55 moves

Block 4 to Position B.

1

Blocks 1 & 2 to Position A.

3

Blocks 3 onto Block 4.

1

60 moves

Extensions
1. 5 moves
2. 57 moves

Chess Queens
There are several different solutions for each of the following situations.

Seven Queens ...

Six Queens ...

Five Queens ...

Dice Differences
There is no specific answer to the game however, the theoretical frequency of the possible differences can guide placement strategy.

Difference

Escape rolls

0

6 ways to escape

1

10 ways to escape

2

8 ways to escape

3

6 ways to escape

4

4 ways to escape

5

2 ways to escape

36 possible rolls

Extension
So far it is not certain which is the best placement strategy. Some possibilities are:

1

2

2

1

0

0

On average this takes about 13.7 rolls

2

2

2

0

0

0

On average this takes about 15.5 rolls

1

3

2

0

0

0

On average this takes about 14.4 rolls

Eric The Sheep
If there are 50 in front of Eric to start with, 17 will be shorn before Eric gets to the front. Remember Eric's intention is to get to the front, so if there aren't two sheep to skip past when the shearer's back is turned, Eric will sneak past the one which is in the way.

A table of results for different numbers of sheep shows an unusual pattern:

Sheep in Front

1

2

3

4

5

6

7

8

9

...

41

50

100

211

Sheep shorn

1

1

1

2

2

2

3

3

3

...

14

17

34

71

One rule is: Divide by 3 and round up if there is a remainder.

Another way to find the answer is expressed symbolically as: INT[(n - 1)/3] + 1

INT means to take the whole number (or integer) part of the an answer. So these symbols mean first subtract 1 from the number of sheep, then divide by 3, then use only the whole number part of the answer, and finally, add 1.

Another symbolic expression which gives the same result is: INT[(n + 2)/3]

Sneaking past three at a time becomes: INT[(n + 3)/4]

Sneaking past x at a time becomes: INT[(n + x)/(x + 1)]

With 2 shearers, and sneaking past 3 we get: 2 x INT[(n - 1)/(2 + 3)]

Eureka
In total there are over 30 solutions. One is:

Peter: (5, 6, 9) = 20

Paul: (7, 8, 10, 15) = 40

Mary: (11, 14, 16, 17, 40) = 98

but: How do we know when we have found them all?

To find all the solutions one could systematically test all combinations for Peter and Paul from the lowest for Peter (5, 6, 7) = 18, to the highest for Paul (15, 16, 17, 40) = 88. Remember Paul's total must be twice that of Peter.

First Down The Mountain
The number of ways each total can be rolled, ie:

Total

2

3

4

5

6

7

8

9

10

11

12

No. of Ways

1

2

3

4

5

6

5

4

3

2

1

seems to have been built into the symmetry of the board. We could reason that every climber has an equal chance of reaching the second last row, then, since Climber 7 has more chance of being rolled, it has more chance of reaching the winning line first.

This intuitive reasoning seems reasonable, but it doesn't hold up when compared to the empirical data gained by playing the game many times. Climbers 2 and 12 win most often. How can this be explained?

In fact, the explanation is not easy, but an indication of the reason comes by considering a race between the 2 and the 7, ignoring all other rolls, where the 2 has to go one step and the 7 has to go six steps.

From the table above, there are only 7 roll results which would be of interest. Six of these are wins for Climber 7 and only one is a win for Climber 2. So, Climber 7 would win 6/7 of the time. Climber 2 can win on any throw, but the only way Climber 7 can win is to roll 6 successive 7s before a single 2 is rolled. The theoretical chance of this is:

6/7 x 6/7 x 6/7 x 6/7 x 6/7 x 6/7 = 39.6%.

So in 1,000 trials we would expect Climber 7 to win 396 times which would mean Climber 2 would win 604 times.

Four In A Row (discontinued)
To date, arrangements of counters have been found to give scores for all values up to 40 except 1, 2, 32, 35, 38 and 39. If you find arrangements for these points please let us know. Alternatively perhaps you could prove that one or more of these can't possibly be made.

Examples

This arrangement for one colour scores 28 points.

This arrangement for one colour scores 40 points.

This is one arrangement for getting zero points.

Garden Beds

8 2. 10

A table of results shows:

Length of Bed (L)

1

2

3

4

5

No. of Tiles (T)

8

10

12

14

16

For any size bed the number of tiles is twice the length of the bed, plus another six. That is:

T = 2 x L + 6

It is twice because you need to surround both sides of the bed, then six extras allows for three at each end. If the bed was 100 units long, you would need 100 on one side, 100 on the other side, and 3 at each end. Therefore if N = 100, T = 206.
There are other ways to 'see' the bed which will help you work out the number of tiles. Can you 'see' things a different way?

Extensions

Length of Bed (L)

23

38

100

247

1,396

40

140

982

5.5

No. of Tiles (T)

52

82

226

500

2,798

86

286

1,970

17

Hearts & Loops
The key to solving this puzzle is in seeing the hoop on the rod pass through the dip of the heart. It is something like a key in a lock image.

Place the central dip of the heart into this loop.

Bring the large hoop(on the end of the rod) through the dip, from position A to position B.

Lift the heart out of the loop to detach the pieces.

Highest Number 2

There are no specific answers to this task.

One analysis of a strategy, supposing the first two rolls were 5 and then 6, could be:
5 is rolled first so place it in the centre position because it is in the middle of 1-9.

5

The next roll is a 6. After that there are 7 numbers left for the third roll. If 6 is placed in the units, four of these (4, 3, 2, 1) would be a poor result, and only three (7, 8, 9) are good. But if the 6 is placed in the hundreds:

Magic Cube
The solution is shown here with the three vertical layers separated. The front one is on the left.
6 is in the top left corner.
14 is in the middle of the central layer.
22 is on the bottom right corner.
Together they make a line from front top left to rear bottom right which has a magic total of 42.

Red To Blue
One approach, for even numbers of counters using the rule of turning all except 1, is:

Count consistently from the left (or right).

On move M, the Mth counter is unchanged and all others turn over.

Continue in order until all counters have been unchanged once and...

Another approach is to draw a diagram like this:

Snail Trail

7 hours

In the 3rd hour (2 hours, 36 minutes)

In the 6th hour (5 hours, 50 minutes, ie: it gets out more quickly.)

(Up 3, Slip 2) or (Up 5, Slip 2) for example.

Extensions

12 metres ... 5 hours

15 metres ... 6 hours (5 hours 30 minutes)

100 metres ... 49 hours (48 hours 24 minutes)

In the 5th hour (4 hours 30 minutes)

The important height (from which it can reach the top in the next hour) is 11 - x.
It gains (x - y)/2 metres per hour.
Hence it reaches 11 - x in (11 - x)/[(x - y)/2] hours, rounded up to the next hour.
Then it reaches the top within the next hour, so the time is:
(11 - x)/[(x - y)/2] - rounded up - plus 1.

Soma Cube 2
There are no specific answers for this task.

Sphinx
Size 2
requires 4 pieces.
Size 3
requires 9 pieces.
Size 4 is made using the Size 2 as a template and substituting a Size 2 for each Size 1 piece.
This requires 16 pieces.

A Size 5 would require 25 pieces.
Size 6 would need 36 pieces - use Size 3 as the unit and build a double Size 3.
A Size N Sphinx would need NxN pieces. This shows the quadratic relationship between length and area.

Tower Of Hanoi

It takes 31 moves to relocate the tower made of 5 pieces.

For various numbers of pieces, the moves are:

No. of Pieces

2

3

4

5

6

7

8

9

10

...

n

No. of Moves

3

7

15

31

63

127

511

511

1023

...

2n - 1

Notice that each answer is 'double the previous answer plus 1'.
- First move the 'top' 3 pieces to another peg (7 moves).
- Then move the base piece to the empty peg (1 move).
- Then move the top 3 pieces back onto the base (7 moves).

Similarly, to move a tower of 10, first move the top 9 pieces (511) moves, then the base piece (1 move), then the top 9 back again (511 moves), for a total of 1023.

The formula m = 2n - 1 works for any number of discs.
If n = 20, m = 1,048,575.
Moving one disc per minute would take 104.02 weeks (or two years).

A tower of 100 pieces would take hundreds of billions of years.
(2,400,000,000,000,000,000,000,000 years)

The most frequent result is 50¢. If you reached the position immediately below the 20¢ prize, the chance of going left or right is 20/36, compared to going ahead (16/36).

The largest prize ($5) is placed where the same rolls are needed one after the other, ie: LLL or RRR. Theoretically, the chance of each of these is:10/36 x 10/36 x 10/36 = 2.14%
Hence 4.28% of players can expect to win $5.

The operator in this setup would LOSE money.
For 1000 players, the money in is $1000.
The payout is $1,263.40 for a LOSS of $263.40 or 26.3%.
If, for example, the $1 prize was changed to NIL, then the operator would just break even.

Extension
There are six pathways to win $4 on either side, hence 12 altogether, eg: RLLL or LULLL.